Answered By Leonardo F

Using the generic term formula for an arithmetic sequence:

 $tn=t1+\left(n-1\right)d$tn=t1+(‍n−1)d 

 $t7=64$t7=64 

 $t26=235$t26=235 

We can write:

 $t26=t7+\left(26-7\right)d$t26=t7+(26−7)d 

 $t26=t7+19d$t26=t7+19d 

 $235=64+19d$235=64+19d 

Solving for the common difference:

 $d=\frac{235-64}{19}=9$d=235−6419 =9 

So we can calculate the first term:

 $t7=t1+6d$t7=t1+6d 

 $t1=t7-6d=64-6\left(9\right)=10$t1=t7−6d=64−6(9)=10 

The 15th term will be:

 $t15=t1+14d=10+14\left(9\right)=136$t15=t1+14d=10+14(9)=136

Finally, the sum of the first 15 terms will be:

 $S15=\frac{\left(t1+t15\right)15}{2}=\frac{\left(10+136\right)15}{2}=1095$S15=(t1+t15)152 =(10+136)152 =1095 

Answered By Sujalakshmy V

tn= a+(n-1)d where a is the first term and d is the common difference.

Here   seventh term is given 

t7= a+(7-1)d=64

t26=a+(26-1)d=235

a+6d=64 ......(1)

a+25d=235....(2)

On solving these simultaneous linear equations  ,we get d= 9.

Then substitute in (1) to get a ,  a=10.

Sum to 15 terms is found using the formula 

Sn=(n/2)(2a+(n-1)d)

S15=(15/2)(2×10+14×9) =1095

The sum of first 15 terms is 1095.

Answered By Mahboubeh D

t_n=t₁+(n-1)d

64=t₁+6d

235=t₁+25d

d=9 and t₁=10

so t₁₅=136

then s₁₅=[(t₁+t₁₅)*15]/2

S₁₅=1095